Q. 2 C4.6( 9 Votes )

# D and E are point

Answer :

Given: AD = (7x – 4) cm, AE = (5x – 2), DB = (3x + 4) cm and EC = 3x cm

By Thale’s theorem,  3x(7x – 4) = (5x – 2)(3x + 4)

21x2 – 12x = 15x2 + 20x – 6x – 8

21x2 – 12x = 15x2 + 14x – 8

21x2 – 15x2 – 12x – 14x + 8 = 0

6x2 – 26x + 8 = 0

2×(3x2 – 13x + 4) = 0 [Simplifying the equation]

3x2 – 13x + 4 = 0

3x2 – 12x – x + 4 = 0

3x(x – 4) – (x – 4) = 0

(3x – 1)(x – 4) = 0

(3x – 1) = 0 or (x – 4) = 0

x = 1/3 or x = 4

Now since we’ve got two values of x, that is, 1/3 and 4. We shall check for its feasibility.

Substitute x = 1/3 in AD = (7x – 4), we get

AD = 7×(1/3) – 4 = -1.67, which is not possible since side of a triangle cannot be negative.

Hence, x = 4 cm.

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